This quiz tests the work covered in Lecture 17 and corresponds to Section 4.1 of the
textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et
al.).

There are more web quizzes at Wiley, select Section 1.

You might want to go back to some of the resources for the second derivative if they were a bit hard at the time. They were http://archives.math.utk.edu/visual.calculus/3/graphing.14/ although some of the language is technical. There are also difficult, but useful quizzes at http://archives.math.utk.edu/visual.calculus/3/graphing.2/index.html and http://archives.math.utk.edu/visual.calculus/3/graphing.3/index.html.

There is some detailed information on this topic with exercises with fully worked solutions at http://www.math.ucdavis.edu/ kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html.

Differentiate $f$ and draw a sign diagram of ${f}^{\prime}\left(t\right)$ to decide which one of the following statements is true. Exactly one option must be correct)

*Choice (a) is correct!*

This shows that ${f}^{\prime}$ changes from positive to negative at $t=-2$ and we can see that means that there is a local maximum at $\left(-2,12\right)$

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

*Choice (a) is incorrect*

*Choice (b) is correct!*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

Find the first and second derivatives of $f$ and decide which of the following statements are true. (Zero or more options can be correct)

*There is at least one mistake.*

For example, choice (a) should be False.

*There is at least one mistake.*

For example, choice (b) should be True.

*There is at least one mistake.*

For example, choice (c) should be True.

*There is at least one mistake.*

For example, choice (d) should be False.

*There is at least one mistake.*

For example, choice (e) should be True.

*There is at least one mistake.*

For example, choice (f) should be False.

*Correct!*

*False*You may have used the function to construct the sign diagram, instead of the derivative.*True*$$\begin{array}{llll}\hfill {f}^{\prime}\left(x\right)& =2x{\left(x-2\right)}^{3}+{x}^{2}\times (3{\left(x-2\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(2\left(x-2\right)+3x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(5x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ This tells us that there are critical points at $x=0,\frac{4}{5}$ and $2$. Construct a sign diagram to determine the nature of these points: Hence there is a stationary points of inflection at $x=2$ and a local minimum at $x=\frac{4}{5}$ and a local maximum at $x=0$*True*$$\begin{array}{llll}\hfill {f}^{\prime}\left(x\right)& ={\left(x-2\right)}^{2}\left(5{x}^{2}-4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {f}^{\u2033}\left(x\right)& =2\left(x-2\right)\left(5{x}^{2}-4x\right)+{\left(x-2\right)}^{2}\left(10x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(x-2\right)\left(2\left(5{x}^{2}-4x\right)+\left(x-2\right)\left(10x-4\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\left(x-2\right)\left(5{x}^{2}-8x+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ Using the quadratic formula to find where $5{x}^{2}-8x+2=0$ we have:$$x=\frac{8\pm \sqrt{64-40}}{2\times 5}=\frac{8\pm \sqrt{24}}{10}=\frac{4\pm \sqrt{6}}{5}.$$Note that $\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}$. So there are points of inflection at $x=2$, $x=\frac{4+\sqrt{6}}{5}$ and $x=\frac{4-\sqrt{6}}{5}$*False*You have made the same mistake as in b).*True*This is true but there are more features.*False*You can factorize the second derivative and get 3 points of inflection.